ARCHIVUM MATHEMATICUM (BRNO) Tomus 40 (2004), 321 – 333
THE NEUMANN PROBLEM FOR QUASILINEAR DIFFERENTIAL EQUATIONS
TIZIANA CARDINALI, NIKOLAOS S. PAPAGEORGIOU AND RAFFAELLA SERVADEI
Abstract. In this note we prove the existence of extremal solutions of the quasilinear Neumann problem−(|x0(t)|p−2x0(t))0 =f(t, x(t), x0(t)), a.e. on T,x0(0) =x0(b) = 0, 2≤p <∞in the order interval [ψ, ϕ], whereψandϕ are respectively a lower and an upper solution of the Neumann problem.
1. Introduction
Recently several authors studied quasilinear ordinary differential equations of second order. People mainly studied the Dirichlet and the periodic problem using a variety of methods. We refer to the works of Boccardo-Dr´abek-Giachetti-Kuˇcera [1], Dr´abek [4], Del Pino-Elgueta-Manasevich [3], Guo [9], O’Regan [13] and Wang- Jiang [16]. Of all these works only Guo in 1993 considers a Neumann problem, but he assumes that his vector fieldsf is independent of the derivative. Moreover, he establishes only the existence of solutions for the Neumann boundary value problem and his approach is different from ours: it is based on degree theoretic techniques. In contrast here we allowf to depend on the derivative and we use the method of upper and lower solutions coupled with the theory of nonlinear operators of monotone type. Assuming the existence of an upper solution ϕand a lower solution ψ such that ψ ≤ ϕ, we obtain the existence of a solution, be- longing toC1(T), of the quasilinear Neumann problem in the order interval [ψ, ϕ]
as a consequence of a fixed point result called the nonlinear alternative of Leray- Schauder. Then we also show the existence of extremal solutions in [ψ, ϕ], i.e. of solutionsx∗ andx∗in [ψ, ϕ] such that any other solutionxin [ψ, ϕ] of the bound- ary value problem satisfiesx∗ ≤x ≤x∗. The upper and lower solutions method was also used by Wang-Jiang in [16] and, in the context of semilinear second order periodic boundary value problems, by Gao-Wang in [6] and, for multivalued semi- linear Sturm-Liouville problems, by Halidias-Papageorgiou in [10]. The existence
2000Mathematics Subject Classification: 35J60, 35J65.
Key words and phrases: upper solution, lower solution, order interval, truncation function, penalty function, pseudomonotone operator, coercive operator, Leray-Schauder principle, maxi- mal solution, minimal solution.
Received October 14, 2002.
of extremal solutions in [ψ, ϕ] was addressed only by Halidias-Papageorgiou, using different approach.
2. Preliminaries
If T = [0, b], let W1,p(T) = {x ∈ Lp(T) | x0 ∈ Lp(T)}be the function space equipped with the normkxk= (kxkpp+kx0kpp)1p: the spaceW1,p(T) is a separable, reflexive Banach space for all 1< p <∞. It is well known that W1,p(T) embeds continuously intoC(T), i.e. every element inW1,p(T) has a unique representative intoC(T). MoreoverW1,p(T),→Lp(T) compactly for all 1≤p <∞.
LetX a reflexive Banach space andX∗ its topological dual. In what follows by (., .) we denote the duality brackets of the pair (X, X∗). A mapA:X →X∗is said to be ‘monotone’, if for allx1, x2∈X, we have (A(x2)−A(x1), x2−x1)≥0. A(·) is said to be ‘strictly monotone’ if it is monotone and (A(x2)−A(x1), x2−x1) = 0 impliesx2 =x1. We say thatA(·) is maximal monotone, if its graph is maximal monotone with respect to inclusion among the graphs of all monotone maps fromX intoX∗. It follows from this definition thatA(·) is maximal monotone if and only ifAis monotone and (v∗−A(x), v−x)≥0 for allx∈X, impliesv∗=A(v). An operatorA:X→X∗is said to be ‘demicontinuous’ atx∈X, if for every{xn}n∈N
with xn →x in X, we haveA(xn)w→∗A(x) in X∗. A monotone demicontinuous everywhere defined operator is maximal monotone (see Hu-Papageorgiou [11]). A map A:X →X∗ is said to be ‘pseudomonotone’, if for allx ∈X and for every sequence{xn}n∈N⊆Xsuch thatxn w
→xinX and lim sup(A(xn), xn−x)≤0, we have that (A(x), x−y)≤lim inf(A(xn), xn−y) for ally∈X. A mapA:X →X∗ is said to be ‘coercive’ if (A(x),x)kxk → ∞as kxk → ∞. A pseudomonotone map which is also coercive is surjective. A single valued operatorA:X →X∗ is said to be ‘compact’ if it is continuous and maps bounded sets into relatively compact sets.
3. Existence result
LetT = [0, b]⊂R. The problem under consideration is the following:
−(|x0(t)|p−2x0(t))0 =f(t, x(t), x0(t)) a.e. onT x0(0) =x0(b) = 0 2≤p <∞
(1)
Let us start by introducing the hypotheses on the right hand side function f(t, x, y).
H(f ): f :T×R×R→Ris a function such that (i) for all (x, y)∈R×R,t7→f(t, x, y) is measurable;
(ii) for almost allt∈T, (x, y)7→f(t, x, y) is continuous;
(iii) for almost all t∈T, allx∈[ψ(t), ϕ(t)] and ally∈R, we have
|f(t, x, y)| ≤a(t) +c|y|p−1 witha∈Lq(T),c >0 and 1p +1q = 1.
Definition 1. By a solution of (1) we mean a function x ∈ C1(T) such that
|x0(·)|p−2x0(·)∈W1,q(T) and it satisfies (1).
As we already mentioned we shall employ the method of upper and lower solu- tions. So let us introduce these two concepts:
Definition 2. By an ’upper solution’ of (1) we mean a functionϕ∈C1(T) such that|ϕ0(·)|p−2ϕ0(·)∈W1,q(T) and it satisfies
−(|ϕ0(t)|p−2ϕ0(t))0 ≥f(t, ϕ(t), ϕ0(t)) a.e. onT ϕ0(0)≤0≤ϕ0(b)
(2)
Definition 3. By a ’lower solution’ of (1) we mean a function ψ ∈ C1(T) such that|ψ0(·)|p−2ψ0(·)∈W1,q(T) and it satisfies
−(|ψ0(t)|p−2ψ0(t))0 ≤f(t, ψ(t), ψ0(t)) a.e. onT ψ0(b)≤0≤ψ0(0)
(3)
We will assume the existence of an upper solutionϕ and a lower solution ψ.
More precisely we make the following hypothesis:
H000: There exist an upper solutionϕ and a lower solutionψ of problem (1) such thatψ(t)≤ϕ(t) for all t∈T.
We introduce the operatorA:W1,p(T)→W1,p(T)∗ defined by hA(x), yi=
Z b 0
|x0(t)|p−2x0(t)y0(t)dt for ally∈W1,p(T).
Proposition 1. A:W1,p(T)→W1,p(T)∗ is maximal monotone.
Proof. It suffices to show that Ais monotone, demicontinuous.
First we show the monotonicity ofA. For allx, y∈W1,p(T) we have hA(x)−A(y), x−yi=
Rb
0(|x0(t)|p−2x0(t)− |y0(t)|p−2y0(t))(x0(t)−y0(t))dt≥0 from the well-known inequality (see Hu-Papageorgiou [11], p. 303).
Next we show the demicontinuity ofA. To this end, letxn →x inW1,p(T) as n→ ∞. By passing to a subsequence if necessary, we may assume thatx0n(t)→ x0(t) a.e. onT and there existsk∈Lp(T) such that |x0n(t)| ≤k(t) a.e. on T for alln ∈N. Thus we deduce that|x0n|p−2x0n → |x0|p−2x0 in Lq(T) asn → ∞. So for everyy∈W1,p(T) we have
hA(xn)−A(x), yi= Z b
0
(|x0n(t)|p−2x0n(t)− |x0(t)|p−2x0(t))y0(t)dt→0
asn→ ∞.
We conclude that A(xn)→w A(x) inW1,p(T)∗, i.e.Ais demicontinuous.
BecauseAis monotone and demicontinuous,Ais maximal monotone.
Let A1 : D1 ⊆ Lp(T) → Lq(T) be defined by A1(x) = A(x) for all x ∈ D1= {y∈W1,p(T)|A(y)∈Lq(T)}.
Proposition 2. A1:D1⊆Lp(T)→Lq(T)is maximal monotone.
Proof. LetJ :Lp(T)→Lq(T) be defined byJ(x)(·) =|x(·)|p−2x(·). To establish the result of the proposition, it suffices to show thatR(A1+J) =Lq(T). Indeed, let this surjectivity condition be true and letv∈D1,v∗∈Lq(T) be such that
(A1(x)−v∗, x−v)pq≥0 (4)
for all x ∈ D1. Since we have assumed that R(A1+J) = Lq(T), we can find x0∈D1 such that
A1(x0) +J(x0) =v∗+J(v) Using this in (4) above, we obtain
(J(v)−J(x0), v−x0)pq≥0.
ButJ(·) is strictly monotone. Hence it follows thatx0 =v and soA1(x0) = v∗. This proves the maximality of A1(·). To show the previous range condition, we proceed as follows. Let ˆJ =i◦J|W1,p(T) where i is the embedding operator and J|W1,p(T) is the restriction of J on W1,p(T). Then ˆJ : W1,p(T) → W1,p(T)∗ is continuous, monotone, D( ˆJ) = W1,p(T), hence maximal monotone (see Hu- Papageorgiou [11], Corollary III.1.35, p. 309). So from this fact together with Proposition 1 and Theorem III.3.3, p. 334 of Hu-Papageorgiou [11], we have that A+ ˆJ :W1,p(T)→W1,p(T)∗ is maximal monotone. Moreover, we have
hA(x) + ˆJ(x), xi=hA(x), xi+ ( ˆJ(x), x)pq =kx0kpp+kxkpp =kxkp1,p. ThusA+ ˆJ(·) is coercive. But a maximal monotone coercive operator is surjective (see, for example, Corollary III.2.19, p. 332 of Hu-Papageorgiou [11]). Hence R(A+ ˆJ) =W1,p(T)∗. Therefore giveng∈Lq(T) we can findx∈W1,p(T) such that A(x) + ˆJ(x) = g. So A(x) = g−Jˆ(x) ∈ Lq(T) and then A(x) = A1(x).
Hence A1(x) +J(x) = g. Because g ∈ Lq(T) was arbitrary, we conclude that R(A1+J) =Lq(T).
In the next proposition we describe the range of A1 by means of a boundary value problem.
Proposition 3. If g ∈ R(A1) then, for every x ∈ D1 such that A1(x) = g, we have
−(|x0(t)|p−2x0(t))0 =g(t) a.e. onT x0(0) =x0(b) = 0
(5)
andx∈C1(T)with|x0(·)|p−2x0(·)∈W1,q(T).
Proof. For everyϕ∈C0∞((0, b)) we have
(A1(x), ϕ)pq= (g, ϕ)pq
that is
Z b 0
|x0(t)|p−2x0(t)ϕ0(t)dt= Z b
0
g(t)ϕ(t)dt (6)
So from the definition of the distributional derivative we infer that
−(|x0|p−2x0)0 =g and|x0|p−2x0 ∈W1,q(T).
Recall thatW1,q(T) is embedded continuously inC(T). So|x0(·)|p−2x0(·)∈ C(T).
Leth:R→Rbe defined byh(r) =|r|p−2r. This is continuous, strictly monotone and soh−1 :R→Ris well-defined and it is easily seen to be continuous. Hence h−1(|x0(·)|p−2x0(·)) =x0(·)∈C(T). Using Green’s formula (integration by parts), for everyy∈W1,p(T), we have
(7) Z b
0
g(t)y(t)dt=− Z b
0
(|x0(t)|p−2x0(t))0y(t)dt
= Z b
0
|x0(t)|p−2x0(t)y0(t)dt − |x0(b)|p−2x0(b)y(b) +|x0(0)|p−2x0(0)y(0). So, by equality (6), we have
|x0(b)|p−2x0(b)y(b) = |x0(0)|p−2x0(0)y(0). (8)
Lety ∈W1,p(T) be such that y(0) =y(b) = 1 (for example takey(t) = 1 for all t∈T). We have
|x0(b)|p−2x0(b) = |x0(0)|p−2x0(0). Hence
h−1(|x0(b)|p−2x0(b)) =h−1(|x0(0)|p−2x0(0))
and sox0(b) =x0(0). Using this in (8) and becausey ∈W1,p(T) is arbitrary, we conclude thatx0(b) =x0(0) = 0.
Our application of the upper and lower solutions method will proceed via truncation and penalization techniques. So we introduce the truncation map τ : W1,p(T)→W1,p(T) be defined by
τ(x)(t) =
ϕ(t) if ϕ(t)≤x(t) x(t) if ψ(t)≤x(t)≤ϕ(t) ψ(t) if x(t)≤ψ(t).
It is clear thatτ(·) is continuous.
We observe thatτ(x)0 ∈Lp(T) being τ(x)0(t) =
ϕ0(t) if ϕ(t)≤x(t) x0(t) if ψ(t)≤x(t)≤ϕ(t) ψ0(t) if x(t)≤ψ(t). The penalty functionu:T×R→Ris defined by
u(t, x) =
(x−ϕ(t))p−1 if ϕ(t)≤x
0 if ψ(t)≤x≤ϕ(t)
−(ψ(t)−x)p−1 if x≤ψ(t). It is clear thatu(·,·) is a Carath´eodory function such that
|u(t, x)| ≤a1(t) +c1|x|p−1 a.e. on T and
Z b 0
u(t, x(t))x(t)dt≥ kxkpp−c2kxkp−1p for all x∈Lp(T)
witha1∈Lq(T) andc1, c2>0. Usingτ(·) andu(·,·), we introduce the following auxiliary problem:
−(|x0(t)|p−2x0(t))0 =f(t, τ(x)(t), τ(x)0(t))−α u(t, x(t)) a.e. onT
x0(0) =x0(b) = 0 2≤p <∞
(9)
where α > 0. In what follows let K = [ψ, ϕ] = {x ∈ C1(T) | ψ(t) ≤ x(t) ≤ ϕ(t) for allt∈T}.
Proposition 4. If hypothesesH0andH(f)hold, then problem (1)has a solution x inK= [ψ, ϕ].
Proof. First we establish the existence of solutions for problem (9) and then we show that every such solution belongs in K. Hence from the definition of the truncation map τ(·) and the penalty function u(·,·), we can conclude that these are also solutions of problem (1). From the proof of Proposition 2 we know that G = A1 +J : D1 ⊆ Lp(T) → Lq(T) is surjective and strictly monotone. So G−1:Lq(T)→D1 is well defined.
Claim 1: G−1 is strongly continuous from Lq(T) to W1,p(T) (see Zeidler [17], p.597; remark that strong continuity implies compactness and strong continuity is also referred to as complete continuity).
We need to show that ifyn→w yinLq(T), thenG−1(yn)→G−1(y) inW1,p(T).
Letxn =G−1(yn),n∈Nandx=G−1(y). We have A1(xn) +J(xn) =yn. and so
(A1(xn), xn)pq+ (J(xn), xn)pq= (yn, xn)pq.
By definition ofA1 andJ, we obtain
kx0nkpp+kxnkpp ≤M1kxnkp
wherekynkq≤M1for alln∈N. Then we conclude that kxnk1,p≤M2
for someM2>0 and alln∈N. Thus, by passing to a subsequence if necessary, we may assumexn→w xˆinW1,p(T) andxn→ˆxinLp(T) (sinceW1,p(T) is embedded compactly inLp(T)). So we have
(A1(xn) +J(xn), xn−x)ˆpq= (yn, xn−x)ˆpq→0
as n→ ∞. SinceJ(xn)→J(ˆx) inLq(T), beingJ :Lp(T)→Lq(T) continuous, we have
lim sup(A1(xn), xn−x)ˆ pq= lim suphA(xn), xn−xiˆ = 0.
But from Proposition 1 we know that A(·) is maximal monotone and D(A) = W1,p(T), hence A(·) is pseudomonotone. So we have
hA(xn), xni → hA(ˆx),xiˆ and so
kx0nkp→ kˆx0kp.
Since we also have that x0n →w xˆ0 in Lp(T) and the space Lp(T) has the Kadec- Klee property (being uniformly convex), we have thatx0n→xˆ0 in Lp(T) and then xn→xˆinW1,p(T). So we have
A(xn) +J(xn)→w A(ˆx) +J(ˆx) inW1,p(T)∗
asn→ ∞. HenceA(ˆx)+J(ˆx) =yand thenA1(ˆx)+J(ˆx) =y. Because (A1+J)(·) is strictly monotone, we infer that ˆx=x. Therefore we conclude thatxn →x in W1,p(T), which proves the claim.
Next letHα:W1,p(T)→Lq(T) be defined by
Hα(x)(·) =f(·, τ(x)(·), τ(x)0(·))−α u(·, x(·)) +J(x)(·).
By virtue of hypothesesH(f), the properties ofτ(·) andu(·,·), we infer thatHα is bounded and continuous.
Now using Proposition 3, we see that the solvability of problem (9) is equivalent to solving the fixed point problem
x=G−1Hα(x). (10)
SinceG−1Hαis compact, to solve problem (10), we shall use the Leray-Schauder principle. This require to show that the set
Sˆα={x∈W1,p(T)|x=λG−1Hα(x) for some 0< λ <1}
is bounded.
Claim 2: ˆSα⊆W1,p(T) is bounded for someα >0.
Ifx∈Sˆα, we have
G(x
λ) =Hα(x) for someλ∈]0,1[ and so
(G(x
λ), x)pq= (Hα(x), x)pq. By definition ofGand Hα, we obtain
Z b 0
x0(t) λ
p−2x0(t)
λ x0(t) dt + Z b
0
x(t) λ
p−2x(t)
λ x(t) dt
= Z b
0
f(t, τ(x)(t), τ(x)0(t))x(t)dt
−α Z b
0
u(t, x(t))x(t)dt+ Z b
0
|x(t)|p−2x(t)x(t)dt.
Using hypothesesH(f)(iii), the properties ofτ(·) andu(·,·) and Young’s inequal- ity, we have
1
λp−1(kx0kpp+kxkpp)
≤ kakqkxkp+ ˜ckxkp+ ¯ckx0kp−1p kxkp− αkxkpp+α c2kxkp−1p +kxkpp
≤(kakq+ ˜c)kxkp+ ¯c
qqkx0kpp + ¯cp
p −α+ 1
kxkpp+ α c2kxkp−1p
≤(kakq+ ˜c)kxkw+ ¯c
qqkxkpw + ¯cp
p −α+ 1
kxkpp+ α c2kxkp−1w where ˜c, c, c¯ 2, >0. So we obtain
1 λp−1 − ¯c
kxkpw≤(kakq+ ˜c)kxkw+¯cp
p −α+ 1
kxkpp+α c2kxkp−1w (11)
Now, choosing >0 such that 1 λp−1 − ¯c
qq >0 andα >0 such that
α > ¯cp p + 1,
we can conclude, by using the inequality (11), that ˆSαis bounded inW1,p(T) and this proves Claim 2.
Claims 1 and 2 allow us to apply the Leray-Schauder principle and obtain x∈D1 such thatx=G−1Hα(x). ThenG(x) =Hα(x) and soA1(x) = ˆHα(x)∈ Lq(T), where ˆHα(x)(·) =f(·, τ(x)(·), τ(x)0(·))−α u(·, x(·)). So by virtue of Propo- sition 3, we have thatxis a solution of the auxiliary problem (9), withα > ¯cpp+1, where ¯c >0 and >0 are such that λp−11 −q¯cq >0.
Next we show thatx∈K= [ψ, ϕ]. Fixedαas above, we have A1(x) = ˆHα(x).
and so
(A1(x),(ψ−x)+)pq= ( ˆHα(x),(ψ−x)+)pq. By definition ofA1 and ˆHα, we obtain
(12) Z b
0
|x0(t)|p−2x0(t)((ψ−x)+)0(t)dt
= Z b
0
f(t, τ(x)(t), τ(x)0(t))(ψ−x)+(t)dt−α Z b
0
u(t, x(t))(ψ−x)+(t)dt . Also sinceψ(·) is a lower solution, by definition we have
Z b 0
|ψ0(t)|p−2ψ0(t)((ψ−x)+)0(t)dt≤ Z b
0
f(t, ψ(t), ψ0(t))(ψ−x)+(t)dt . (13)
From (12) and (13), we obtain Z
{ψ≥x}
(|x0(t)|p−2x0(t)− |ψ0(t)|p−2ψ0(t))(ψ−x)0(t)dt
≥ Z b
0
(f(t, τ(x)(t), τ(x)0(t))−f(t, ψ(t), ψ0(t)))(ψ−x)+(t)dt (14)
−α Z b
0
u(t, x(t))(ψ−x)+(t)dt . Note that
Z
{ψ≥x}
(|x0(t)|p−2x0(t)− |ψ0(t)|p−2ψ0(t))(ψ−x)0(t)dt≤0 (15)
and
Z b 0
(f(t, τ(x)(t), τ(x)0(t))−f(t, ψ(t), ψ0(t)))(ψ−x)+(t)dt
= Z
{ψ≥x}
(f(t, ψ(t), ψ0(t))−f(t, ψ(t), ψ0(t)))(ψ−x)(t)dt= 0. (16)
Using (15) and (16) in (14) we obtain
−α Z b
0
u(t, x(t))(ψ−x)+(t)dt≤0. By definition of the penalty function, we have
Z b 0
(ψ−x)p−1(t)(ψ−x)+(t)dt≤0 and so
k(ψ−x)+kpp ≤0.
Then sinceψ−x∈C(T), we haveψ(t)≤x(t) for allt∈T.
In a similar fashion we show thatx ≤ϕ. Sox∈K = [ψ, ϕ]. Thereforex is a solution of problem (1) inK.
Next we establish the existence of extremal solutions in the order intervalK= [ψ, ϕ]
Theorem 1. If hypotheses H0 andH(f)hold then problem (1) has extremal so- lutions inK.
Proof. LetSbe the set of solutions of problem (1) in the order intervalK= [ψ, ϕ].
We show that (S,≤) is directed, i.e. for allx1, x2∈S there existsx3∈Ssuch that x1≤x3andx2≤x3. For this purpose, fixedx1, x2∈S, we consider the function y=x1∨x2.
For each >0, let ξ∈Lip(R) be defined by
ξ(t) =
0 if t≤0
t
if 0< t <
1 if ≤t .
Note thatξ→χ{t>0} as↓0. Letθ∈C0∞((0, b)), θ≥0 and set θ1= (1−ξ(x2−x1))θ and θ2=ξ(x2−x1)θ .
Evidentlyθ1, θ2≥0 and, from the chain rule for Sobolev functions, we have (θ1)0 =θ0−ξ0(x2−x1)(x2−x1)0θ−ξ(x2−x1)θ0
and
(θ2)0 =ξ0(x2−x1)(x2−x1)0θ+ξ(x2−x1)θ0. Since by hypothesisx1, x2∈S we have
(A1(x1), θ1)pq = ( ˆf(x1), θ1)pq
and
(A1(x2), θ2)pq = ( ˆf(x2), θ2)pq
with ˆf(y(·)) =f(·, y(·), y0(·)) for all y∈W1,p(T). We have (A1(x1), θ1)pq+ (A1(x2), θ2)pq
= Z b
0
(|x02(t)|p−2x02(t)− |x01(t)|p−2x01(t))ξ0(x2−x1)(t)(x2−x1)0(t)θ(t)dt +
Z b 0
(|x02(t)|p−2x02(t)− |x01(t)|p−2x01(t))ξ(x2−x1)(t)θ0(t)dt +
Z b 0
(|x01(t)|p−2x01(t)θ0(t)dt . Note thatξ0(x2−x1)θ≥0 and
(|x02(t)|p−2x02(t)− |x01(t)|p−2x01(t))(x2−x1)0(t)≥0.
Hence
(A1(x1), θ1)pq+ (A1(x2), θ2)pq
≥ Z b
0
(|x02(t)|p−2x02(t)− |x01(t)|p−2x01(t))ξ(x2−x1)(t)θ0(t)dt +
Z b 0
(|x01(t)|p−2x01(t)θ0(t)dt . Passing to the limit as↓0, we obtain
lim inf[(A1(x1), θ1)pq+ (A1(x2), θ2)pq]
≥ Z b
0
|x02(t)|p−2x02(t)χ{x2≥x1}(t)θ0(t)dt +
Z b 0
|x01(t)|p−2x01(t)χ{x2≤x1}(t)θ0(t)dt
= Z b
0
|y0(t)|p−2y0(t)θ0(t)dt= (A1(y), θ)pq. Also we have
( ˆf(x1), θ1)pq+ ( ˆf(x2), θ2)pq→( ˆf(y), θ)pq
as↓0. Thus finally in the limit as↓0, we obtain (A1(y), θ)pq≤( ˆf(y), θ)pq, for allθ∈C0∞((0, b)), θ≥0. Then
Z b 0
|y0(t)|p−2y0(t)θ0(t)dt≤ Z b
0
f(t, y(t), y0(t))θ(t)dt and so, by Green’s identity, sincey0(0) =y0(b) = 0, we have
Z b 0
−(|y0(t)|p−2y0(t))0θ(t)dt≤ Z b
0
f(t, y(t), y0(t))θ(t)dt .
Then, sinceθ∈C0∞((0, b)), θ≥0 is arbitrary, we infer that the functiony=x1∨x2
satisfies the following properties
−(|y0(t)|p−2y0(t))0 ≤f(t, y(t), y0(t)) a.e. onT y0(b) = 0 =y0(0)
(17)
and y is a continuous function such that |y0(·)|p−2y0(·) ∈ W1,q(T). Therefore (we observe that in order to prove the existence of a function x3 is not necessary y ∈ C1(T)), as in the proof of Proposition 4, we can find x3 ∈ S such that y≤x3≤ϕ. Hence (S,≤) is directed.
Next let C be a chain of S. Let x = supC. Using Corollary 7, p. 336 of Dunford-Schwartz [5], we can find{xn}n∈N⊆C non decreasing such thatxn →x inLp(T). We have
A(xn) = ˆf(xn)
and so
(A(xn), xn)pq= ( ˆf(xn), xn)pq. Using hypothesisH(f)(iii), we obtain
kx0nkpp ≤M(kakq+ ckx0nkp−1p )
where M > 0. Then {xn}n∈N ⊆ W1,p(T) is bounded. So we may assume that xn w
→x inW1,p(T). Also we have
lim(A(xn), xn)pq= lim( ˆf(xn), xn)pq= 0 and so, beingApseudomonotone, we can say
kx0nkp→ kx0kp.
Hence, by the Kadec-Klee property,xn→xinW1,p(T). So in the limit asn→ ∞, we haveA(x) = ˆf(x) a.e. onT and so, by Proposition 3,
−(|x0(t)|p−2x0(t))0 =f(t, x(t), x0(t)) a.e. onT x0(0) =x0(b) = 0
(18)
i.e.x∈S.
Apply Zorn’s lemma to obtain a maximal element x∗ of (S,≤). Since (S,≤) is directed,x∗ is unique and it is the maximal solution of problem (1) inK. On the other hand,S is directed with respect to≥, i.e. for allx1, x2∈S there exists x3 ∈ S such that x1 ≥ x3 and x2 ≥ x3 (see Peressini [15], p.3). Similarly we prove the existence of a maximal elementx∗ of (S,≥). Thereforex∗ is the unique minimal solution of problem (1) inK.
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University of Perugia, Department of Mathematics and Computer Science via Vanvitelli 1, Perugia 06123, Italy
National Technical University, Department of Mathematics Zografou Campus, Athens 15780, Greece
e-mail: npapg@math.ntua.gr
University of Roma ‘Tor Vergata’, Department of Mathematics via della Ricerca Scientifica, Roma 00133, Italy
